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never [62]
3 years ago
7

A trade magazine routinely checks the​ drive-through service times of​ fast-food restaurants. Upper AA 9595​% confidence interva

l that results from examining 773773 customers in one​ fast-food chain's​ drive-through has a lower bound of 169.0169.0 seconds and an upper bound of 172.2172.2 seconds. What does this​ mean?
Mathematics
1 answer:
Butoxors [25]3 years ago
6 0

Answer:

One can be 95% confident that the drive-through service times of​ fast-food chain is between 169 seconds and 172 seconds.

Step-by-step explanation:

The confidence interval gives a range of value for the population mean or parameter, which is calculated from the statistic value of the data (that is sample statistic). This range gives the interval which is associated with a certain level of confidence for which the interval will contain the true value of the unknown parameter (true value). In the scenario above, the associated confidence level is 95%.Hence, we can be 95% confident that the true value will be continued with the interval (169 ; 172)

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Answer:

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Step-by-step explanation:

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Can anyone help me integrate :
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Rewrite the second factor in the numerator as

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\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
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Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
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\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
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\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
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Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

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